System Programing Lab

System Programming (9169) Experiment No. 01

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION ♦

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LIST OF EXPERIMENTS AND RECORD OF PROGRESSIVE ASSESSMENT

Sr. No.

Name of experiment Page No.

Date of performance

Date of submission

Assessment Max.10 marks

Sign. of teacher and remark

1 To understand different classes of system softwares

9

2

To study the internal

structure of computer

system and its

assembly language (Model IBM 360/370

15

3

To understand table processing techniques using searching methods

23

4 To understand table processing techniques using sorting methods

29

5 To understand assembler design and implementation steps

35

6

To understand the design and implementation of various phases of compiler.

45

7 To understand the design of loader and linkage editor

53

8 To understand macros and basic design of macro processor

59

Total marks …….. …… Average marks out of 10 ……*

* To be transferred to proforma of CIAAN-2006 (Proforma A-1/A-2) Note : The curriculum of this subject is referred and the above list of experiments is finalized to achieve the desired objectives.



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EXPERIMENT No. 1 1.0 Title:

To understand different classes of Computer System Software

2.0 Prior concepts:

• Computer System organisation

• Hardware / software components

• programming language processors

• Assembly Language programming

• Operating system environment.

3.0 New concepts:

Proposition 1 : Computer Software It is a general term used to describe a collection of computer programs, procedures and documentation that perform some tasks on a computer system. It is divided into three classes: 1. System software. 2. Programming software 3. Application software. Concept structure

• System software: It is any computer software which manages and controls computer hardware so that application software can perform a task. System software performs tasks like transferring data from memory to disk ,disk to memory, or

rendering text onto a display device. Examples: 1. Compilers /Interpreters 2. Assemblers 3. Loaders/Linker 4. Text editors 5. Debuggers 6. Device drivers 7. Operating systems. etc

Two major categories of system softwares are 1. Machine Dependent System Software

• These softwares rely on Internal physical architecture of machine (Machine code, Instruction format ,Addressing mode , Registers.) 2. Machine Independent System Software

•These softwares do not rely on internal physical architecture of machine (Machine code, Instruction format Addressing mode, Registers)

• Programming software: It is usually provides tools to assist a programmer in writing computer programs and software using different programming languages in a more convenient way.

System software Application software

Computer system software

Programming software



2

01010101

11111001

01001000

Examples: Different Kinds of the tools include:

1. End-User Programming:

� Word, Excel, Paint, Chat, Explorer etc

2. Utilities and Analysis Programming: � BASIC ,Visual Basic, C,C++,PASCAL,COBOL,FORTRAN, JavaScript , HTML, etc

• Application software: It is a set of programs comprising various modules that are written for specific computer application.

Examples � Banking, Inventory control ,Payroll , Engineering, Education, GIS etc

Proposition 2 : System software System software consists of a variety of programs that support the operation of a computer and enables the end-user to perform specific, productive tasks

Concept structure Main goal of system software is to reduce the communication gap between humans and computers. System Software performs verity of functions:

• File editing

• Resource Accounting

• I/O management

• Storage management

• Complete various stages of its processing by computer system like – translation relocation, linking, editing, eventual execution

. Proposition 3: System Programming and its components System Program: It is a program which aids in effective execution of general user’s computational requirements on a computer system. Systems programming: It is the activity of designing and implementing the system

programs.

Foundation system programming

End user

Utilities & Analysis

Applications Systems

People

Application Programming

Compiler Assembler Macro Processor

Loader Text Editor Debugging aids Searching & Sorting

I/O Programs

File System

Scheduler

Libraries

Memory Management

Device Management



3

Components of computer system programming:

� Operating system: It is a system software that provides easy interface for user to

communicate with computer system. It also manages all the resources of computer

system.

� Language processor: It is a system software used to convert / translates one form

of language into another form. A language processor may fall into any one of the

following categories.

i. Translator

ii. Assembler

iii. Compiler

iv. Interpreter

Translator: It is a general term used for converting one form of computer language

into another form. The input text of the translator is called as source program and

output text is called as object program. The output of translator needs not always a

machine code. Compiler: It is a system software that accepts a source program “In a high-level language” and produces a corresponding object program. Interpreter: It is a system software that accepts a source program “In a high-level language” and produces a corresponding object program .The major difference between compiler and interpreter are i) Translates source program line by line. ii) Detects one error at a time. Assembler: It is a system software that accepts a source program in assembly language and produces a corresponding object code Macro Processor: A Macro call is an abbreviation (or name) for some code. A macro definition is a sequence of code that has name (macro call). Macro Processor is a system software which performs macro expansion (in assembly language) for every occurrence of macro call.

� Loader: It is a software which accepts the object programs, prepares these programs for execution and loads them into core memory . � Linker: It is a system software that links various subroutines to the main program with respect to memory addresses. � Text Editor: It is a system software that creates and manages text files.

Physical Machine

Operating Systems

Language Processors

Linkers Loaders Text Editors

System Softwares

Macro Processors

Assemblers Interpreters Compilers Translators



4

4.0 Diagram: The structure explaining steps between the HLL to executable program and role of System Programming in its working sequence.

5.0 Learning Objectives: Intellectual skills • To identify different System Software / Programs • To identify System Components: hardware, software, programs Motor skills

• Ability to handle different System Software. 6.0 Conclusion: • To control hardware system while offering services to its user is the function of system software / software system / application software ..........................................

.......................................................................................................................................

.......................................................................................................................................

• .............................................. type of programmer must know the internal (hardware) structure of computer system .

7.0 Laboratory practice: 1. Create, delete, rename, copy the files/directories using a. MS-DOS b. MS Windows c. Linux 2. Develop any two programs in C/C++ and VB and debug, compile, execute the same programs. 3. Write assembly language program for addition of two eight bit numbers and execute it using a. MASM software b. 8086 assembler kit

8.0 Questions: 1. What is system software? 2. List out different system software? 3. What do you mean by programming?



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4. List all kinds of software in chronological manner. 5. Why should you study the system programming? 6. What is the difference between system software and application software? 7. List the name of operating system, language processors, utility programs in your lab. Write their features. 8. Whether BIOS program and Bootstrap loader are system software? Justify your answer 9. Select from given options which one of the host language uses to develop Compiler or interpreter. a. C/C++ b. visual basic c. JAVA d. FORTRAN e. COBOL f. PASCAL 10. Write correct sequence of the following process in program executions. a. compiling b-coding c-linking d- execution/ run e - loading f- debugging

(Space for Answers)



6

(Space for Answers)



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EXPERIMENT No. 2 (Not to be considered for internal assessment)

1.0 Title:

To study the internal structure of computer system and its assembly language (Model IBM 360/370)

2.0 Prior concepts:

• Internal structure of 8085/8086 processor

• Assembly language programming

3.0 New concepts:

Proposition 1 General machine structure of IBM 360/370

• The IBM 360/370 is based on the “stored program computer” concept.

• Major functional blocks: - Instruction interpreter (II) interprets (understand) the intent of instruction fetched

from memory - Location counter( LC) - Stores the address of the instruction to be executed next. - Working register(WR) - Scratch pad registers used for temporary storage - General register - Storage register used by programmer

- Memory address register (MAR) - Contains the address of memory location that is to be read from or write into

- Memory buffer register (MBR) - Contains copy of designated memory location specified by MAR

- Memory controller - Transfers data between MBR and memory according to the

address given by MAR



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Memory Every memory location consists of 8 bit information. Memory address space 224 byte. The addressing on 360 memory consists three components

Registers 16 general purpose registers (32 bit each) The GPRS may be used as base register or index. 4 floating point registers (64 bit each) 1 program status word (64 bit) Data The 360 operates on following data formats

Offset Contents of index register

Absolute Address

Contents of register



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Instructions The 360 instruction formats are as follows The 360 has arithmetic, logical control or transfer and special instructions.



10

Load

-Sto

re-R

egis

ter

Fix

ed

poin

t a

rith

metic

Hexadecimal op code

Mnemonic Meaning (format)

58 48 98 18 12 50 40 90 5A 4A 1A 59 49 19 5D 1D 5C 4C 1C 5B 4B 1B

Load group L LH LM LR LTR Store group ST STH STM Add group A AH AR Compare group C CH CR Divide group D DR Multiply group M MH MR Subtract group S SH SR

Load (RX) Load half word (RX) Load multiple (RS) Load (RR) Load and test (RR) Store (RX) Store half word (RX) Store multiple (RS) Add (RX) Add half word (RX) Add (RR) Compare (RX) Compare half word (RX) Compare (RR) Divide (RX) Divide (RR) Multiply (RX) Multiply half word (RX) Multiply (RR) Subtract (RX) Subtract half word (RX) Subtract (RR)



11

Fix

ed

poin

t a

rith

metic

Tra

nsfe

r

Hexadecimal op code

Mnemonic Compare-group

Meaning (format)

55 D5 95 15 D2 92 54 D4 94 14 56 D6 96 16 57 D7 97 17 8D 89 8C 88 45 05 47 07 46 06

CL CLC CLI CLR Move-group MVC MVI And-group N NC NI NR Or-group O OC OI OR Exclusive-group X XC XI XR Shift SLDL SLL SRDL SRL Linkage group BAL BALR Branch group BC BCR BCT BCTR

Compare logical (RX) Compare logical (SS) Compare logical (SI) Compare logical (RR) Move (SS) Move (SI) Boolean AND (RX) Boolean AND (SS) Boolean AND (SI) Boolean AND (RR) Boolean OR (RX) Boolean OR (SS) Boolean OR (SI) Boolean OR (RR) Exclusive-or (RX) Exclusive-or (SS) Exclusive-or (SI) Exclusive-or (RR) Shift left (double logical) (RS) Shift left (single logical) (RS) Shift right (double logical) (RS) Shift right (single logical) (RS) Branch and link (RX) Branch and link (RR) Branch on condition (RX) Branch on condition (RR) Branch on count (RX) Branch on count (RR)



12

Mis

cella

ne

ou

s

Best for machine

Best for programmer

Proposition 2 : Assembly language of IBM 360/370

An example of assembly language program To write a program that will add 49 to the content of 10 adjacent full words in

memory under the following set of assumptions. a. The 10 numbers to be added are in contiguous locations beginning at absolute location 952. b. The program in core memory is steaming at absolute location 48. c. ‘49’ is at absolute location 948. d. Register 1 contains 48.

Hexadecimal op code

Mnemonic Compare-group

Meaning (format)

9E 41 9C 0A 9D 43 91 42

HIO LA SIO SVC TIO IC TM STC

Halt I/O (RX) Load address (RX) Start I/O (RX) Supervisor call (SI) Test I/O (RX) Insert character (RX) Test under mask (SI) Store character (RX)

Programs Comments

LOOP TEN FOUR FOURTY9 DATA

START BALR 15,0 USING BEGIN+2,15 SR 4,4 L 3,TEN L 2,DATA(4) A 2,FORTY9 ST 2,DATA(4) A 4,FOUR BCT 3,LOOP BR 14 DC F’10’ DC F’4’ DC F’49’ DC F’1,3,3,3,3, 4,5,8,9,0 END

Identifies names of program Start register 15 to address of the next instruction Pesudo-op indicating to assembler register 15 is base register and its content is address of next instruction Clear register 4 (set index=0) Load the number 10 into register 3 Load data (index) into register 2 Add 49 Store updated value of data (index) Add 4 to register 4 (set index = index+4) Decrement register 3 by 1, if result non zero, branch back to loop branch back to caller Constant 10 Constant 4 Constant 49 Words to be processed

High level language

Assembly language

Machine level language



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4.0 Learning Objectives: Intellectual skills • To know the working of registers and functional blocks in IBM 360/370 • To know the memory, data formats, instructions formats and addressing of IBM

360 / 370 Motor skills

• Ability to develop assembly level language programs (IBM 360/370)

5.0 Laboratory practice: 1. Write an IBM 360/370 ALP to add two 8 bit numbers 2. Write an IBM 360/370 ALP to multiply two 8 bit numbers

6.0 Conclusion: • The IBM 360/370 is based on .......................................... concept.

• In RX instruction format, to get a absolute address, the offset is added to contents of ....................... register and .........................register.

• Relative addresses/absolute addresses are the actual addresses of core memory

7.0 Questions:

Write answers to Q.....Q.....Q.....Q.....and Q..... (Teacher shall allot the questions)

1. Define High level language Assembly language Machine language 2. What do you mean by Machine -OP Pseudo-OP Literal Symbol 3. Why IBM 360/370 uses 12 bit offset instead of 24 bit offset? 4. What is the use of BASE register and index register? 5. State the difference between USING and BALR pseudo-OP 6. How 8088’s addressing differs from IBM 360/370? (Space for answers)



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(Space for answers)


MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION ♦ 15

EXPERIMENT No. 3

1.0 Title: To understand the Table Processing Techniques using Searching Techniques.

2.0 Prior concepts: • Data Structure concept (Array, Pointer, Stack, searching and sorting)

• Basic Assembly language 8085/86 or IBM 360/370 .

3.0 New Concepts: Proposition1: Table Processing Technique It is often required to maintain large tables of information in such a way that items may be moved in and out quickly and easily. For example, in assembler symbol table is composed of multiple-word entries in a fixed format. In the table there is, symbol name (key), its value, and re-locatability. It is necessary to maintain large tables of information in such a manner of symbol corresponding location and its value. Table processing technique classified as given below 1. Search techniques 2. Sort techniques

Concept structure

Proposition 2 : Searching Technique

The problems of searching is as follows 1. More than one entry with the same keyword (for defined and undefined symbols) 2. No entry found ( for defined and undefined symbols ) required individual treatment depending on the function of the table searching has two different class

• Linear searching

• Binary searching Linear searching: It is a simple searching technique. This technique uses table in which the items have not been ordered .One way to look for a given keyword is to compare every entry in the table with a given keyword. This procedure is fine for short tables and has the simplicity in processes, but for long tables it can be very slow.

Search Techniques Table Processing Technique

• Linear search

• Binary search

• Interchange sort

• Shell sort

• Bucket sort

• Radix exchange sort

• Address calculation sort

Sort Techniques



Illustration of linear search: Symbol Table

The symbol LO is compared with each symbol in the table sequentially. If the symbol LO is found , the search is successful otherwise not successful. Binary Searching : A more systematic way of searching an ordered table . This technique. uses following steps for searching a keywords from the table. 1-Finds the middle entry ( N/2 or (N+1)/2 ) 2- Start at the middle of the table and compare the middle entry with the keyword to be searched. 3- The keyword may be equal to , greater than or smaller than the item checked. 4-The next action taken for each of these outcomes is as follows

• If equal, the symbol is found.

• If greater, use the top half of the given table as a new table search.

• If smaller, use the bottom half of the table Illustration of Binary search : Consider a table of 15 items , suppose we are searching for item IF. 1. First compare IF with middle item LO and find that IF must be in the top half of the table 2. Second comparison with middle term in this half of the table. 3. Shows IF to be in the second quarter. 4. Third comparison with IW shows IF to be in the third eighth of the table (i.e between item 4 and 6 ) 5. Final comparison made with the item in position 5. 6. A comparison failure on the fourth probe would have reveled that the item did not exists in the table.

Number Symbol 1 TI 2 EX 3 FN 4 RN 5 OR 6 IW 7 LE 8 LO 9 NC

10 OP 11 IF 12 RD 13 FU 14 TE 15 AL

LO



4.0 Learning Objectives : Intellectual skills

• To understand the process of searching an element ( symbol ) from table

• To compare the searching techniques. Motor skills

• Ability to develop the source code for linear search and binary search

• Ability to compile correct logical , syntax errors , and execute program.

5.0 Stepwise procedure: Binary search algorithm : BinarySearch(A[0..N-1], value, low, high)

{

if (high < low)

return -1 // not found

mid = (low + high) / 2

if (A[mid] > value)

return BinarySearch(A, value, low, mid-1)

else if (A[mid] < value)

return BinarySearch(A, value, mid+1, high)

else

return mid // found

}

Optional algorithm for binary search

1. location = -1;

2. while ((more than one item in list) and (haven't yet found

target))

2A. look at the middle item

2B. if (middle item is target)

have found target

else

2C. if (target < middle item)

list = first half of list

2D. else (target > middle item)

list = last half of list

end while

3. if (have found target)

location = position of target in original list

Number Symbol Probe1 Probe2 Probe3 Probe4 1 AL 2 EX 3 FN 4 FU = IF > FU 5 IF = IF = IF 6 IW = IF < IW 7 LE 8 LO = IF < LO 9 NC 10 OP 11 OR 12 RD 13 RN 14 TE 15 TI



4. return location as the result

Linear search algorithm : for i:=N:1:-1 do //Search from N down to 1. (The step is -1) if A[i] = x then QuitLoop i;

next i;

Return(i); //Or otherwise employ the value.

Optional algorithm for linear search

int linearSearch(int a[], int first, int last, int key)

{

// function:

// Searches a[first]..a[last] for key.

// returns: index of the matching element if it finds key,

// otherwise -1.

// parameters:

// a in array of (possibly unsorted) values.

// first, last in lower and upper subscript bounds

// key in value to search for.

// returns:

// index of key, or -1 if key is not in the array.

for (int i=first; i<=last; i++) { if (key == a[i]) {

return i; }

} return -1; // failed to find key

} 6.0 Diagram: Comparison of linear and Binary search technique



7.0 Laboratory practice: 1 - Develop source code for linear search technique. 2 - Develop source code for Binary search technique.

8.0 Conclusion: • Linear search average length of Time is...................(T(lin)=A*N / T(bin)=B*log

2(N) )

• Binary search average length time is........................(T(lin)=A*N / T(bin)=B*log

2(N) )

9.0 Questions:

Write answers to Q.....Q.....Q.....Q.....and Q..... (Teacher shall allot the questions) 1. Define the following terms in one or two sentences. a. Searching b. Sorting c. Hashing or random searching 2. Show the result of each pass for the Binary search . ( 81,52,57,22,95,04,83,96,42,32,48,82)

3. Write assembly sample code for binary search method. 4. Draw the flowchart for linear search technique. 5. Draw the flowchart for Binary search technique

(Space for answer)



(Space for answer)



EXPERIMENT No. 4 1.0 Title: To understand the Table Processing Techniques using sorting techniques.

2.0 Prior concepts:

• Data Structure concept (Array, Pointer, Stack, searching and sorting)

• Basic Assembly language 8085/86 or IBM 360/370

3.0 New Concepts : Proposition1 Sorting techniques It is a technique required to rearrange unordered large tables like Machine-Op table (MOT) and pseudo-Op table (POT ) and symbol table (ST) in sequential order. Table processing technique classified as given below 1. Search techniques 2. Sort techniques

Concept structure 1. Interchange sort : It is also known as bubble sort, sinking sort, sifting sort . This simple sort takes adjacent pairs of items in the table and put them in order as required.

Illustration of Interchange sort : In this method the elements is rearranged by exchanging the two adjacent element

If they are not in order

Unsorted List

1st

pass

2 nd

pass 3 rd pass

4th pass

5 th pass

6 th pass

7 th & final pass

19 13 05 27 01 26 31 16 02 09 11 21

13 05 19 01 26 27 16 02 09 11 21 31

05 13 01 19 26 16 02 09 11 21 27 31

05 01 13 19 16 02 09 11 21 26 27 31

01 05 13 16 02 09 11 19 21 26 27 31

01 05 13 02 09 11 16 19 21 26 27 31

01 05 02 09 11 13 16 19 21 26 27 31

01 02 05 09 11 13 16 19 21 26 27 31

Table Processing Technique

• Linear search

• Binary search

• Interchange sort

• Shell sort

• Bucket sort

• Radix exchange sort

• Address calculation sort

Search Techniques Sort Techniques



2. Shell sort : This sort is similar to the interchange sort in that it moves data items by exchange. A fast comparative sort algorithm. It begins by comparing items a

distance‘d’ apart. The items that are way out of place will be moved rapidly than a simple interchange sort.

Illustration of Shell sort :

* = Exchange ** = Dual Exchange

*** = Triple Exchange.

3. Bucket sort : It is one of the simple distributed sort is called the radix or bucket sort. The sort involve examining the least significant digit of the keyword first and the item is then assigned to a bucket uniquely dependent on the value of the digits.

Unsorted List

pass 1 (d

1=6)

pass 2 (d

2=3)

pass 3 (d

3=2)

pass 4 (d

4=1)

19 13 05 27 01 26 31 16 02 09 11 21

19 13 *02 *09 01 *21 31 16 *05 *27 11 *26

*09 *01 02 *19 **11 *05 *27 **13 *21 *31 *16 26

*02 01 *09 *05 11

**13 **16 *19

***21 *26 *27 *31

*01 *02 *05 *09 11 13 16 19 21 26 27 31

Original Table

First distribution

Merge

Second distribution

Final merge

19 13 05 27 01 26 31 16 02 09 11 21

0 1- 01,31,11,21 2- 02 3- 13 4 5- 05 6- 26,16 7-27 8 9- 19,09 09

01 31 11 21 02 13 05 26 16 27 19 09

0-01,02,05,09 1-11,13,16,19 2-21,26,27 3-31 4- 5- 6- 7- 8- 9-

01 02 05 09 11 13 16 19 21 26 27 31

Separate based on last digit

Separate based on last digit



4. Radix exchange sort : A considerably better distributed sort , its used when keys are expressed ( expressible) in binary . This sorting is accomplished by considering groups with the same (M) first bits and ordering that group with respect to the (M+1)st bits. Scanning down from the top of the group for a one bit and up from the bottom for zero bit . Thus that two process are exchange and sorting continues.

5. Address calculation sort : It is one of the fastest type of sort if enough storage space is available. Sorting is done by transforming the key into an address in the table that “ represents “ the key .

4.0 Learning objectives : Intellectual skills

• To understand the process of sorting by applying different methods

• To compare the sorting method by passes , exchanges, and comparisons

• Understand the efficiency of sorting method

Motor skills

• Ability to Convert the different sorting method process into assembly code

(representative 360/370 instructions / SYMTLB ) 5.0 Stepwise procedure / algorithms:

1 Interchange sort algorithm:

procedure bubbleSort( A : list of sortable items ) defined as:

n := length( A )

do

swapped := false

n := n - 1

for each i in 0 to n do:

if A[ i ] > A[ i + 1 ] then

swap( A[ i ], A[ i + 1 ] )

swapped := true

end if

end for

while swapped

end procedure

2. Shell sort algorithm : void shellSort(int numbers[], int array_size)

{



int i, j, increment, temp;

increment = 3;

while (increment > 0)

{

for (i=0; i < array_size; i++)

{

j = i;

temp = numbers[i];

while ((j >= increment) && (numbers[j-increment] > temp))

{

numbers[j] = numbers[j - increment];

j = j - increment;

}

numbers[j] = temp;

}

if (increment/2 != 0)

increment = increment/2;

else if (increment == 1)

increment = 0;

else

increment = 1;

}

}

3. Bucket sort algorithm :

function bucket-sort(array, n) is buckets ← new array of n empty lists

for i = 0 to (length(array)-1) do

insert array[i] into buckets[msbits(array[i], k)]

for i = 0 to n - 1 do

next-sort(buckets[i])

return the concatenation of buckets[0], ..., buckets[n-1]

6.0 Laboratory practice :

Teachers shall allot one of the sorting technique to each group for implementation .

( in c or c++ )

1. write a source code for an Interchange sort algorithm.

2.wirte a source code for a bucket sort algorithm .

3 write a source code for a shell sort algorithm

7.0 Conclusion: Different kinds of sorting methods with its Average time (Approx) and Extra storage (wasted space)

Sr. No.

Type Average time (Approximate)

Extra storage (Wasted space)

1 Interchange

2 Shell

3 Radix

4 Radix Exchange

5 Address calculation



8.0 Questions:

Write answers to Q.....Q.....Q.....Q.....and Q..... (Teacher shall allot the questions) 1. Show the result of each pass for the following lists using a. Inter change sort b. Shell sort c. Radix sort (SWATI, ALKA, VARSHA, RUPALI, MRUNAL, ROHINI, ZARINA, YAMINI, BABALI, DIVYA ) 2. Teacher shall make the group of 5 students and allot them to implement one of the sorting techniques to each group. ( draw the flow chart ) a. Interchange b. Shell c. Radix exchange d. Address calculation e. Radix 3. Define the following terms in two or three sentences. 1. searching 2. sorting 3. hashing (Space for answers)



(Space for answers)



EXPERIMENT No. 5 1.0 Title:

To understand assembler design and implementation steps

2.0 Prior concepts: • Data Structure.

• Assembly language programming concepts.

3.0 New concepts:

Proposition 1 : Assembler definition , features. The assembler is the system program that translate source code written in assembly language to object code( Machine Language) and other information for loader.

Concept structure

General design procedure : 1. Specify the problem 2. Specify data structures 3. Define format of data structures 4. Specify algorithm 5. Look for modularity (i.e capability of one program to be subdivided into independent programming units ) 6. Repeat 1 through 5 on modules

Assembler Design Features There are two different assembler design features. Each feature has its own method

for design and implement assembler. Following features define assembler design and implementation methods.

• Machine Dependent Assembler Features In this method specific instruction format and addressing modes has PC relative and PC Based . o Specified Instruction formats and addressing modes o Needs Program relocation

• Machine independent Assembler Features o Literals o Symbol defining statements o Expressions o Program blocks o Control Sections and program linking.

• Convert mnemonic operation codes to their machine language equivalents. • Convert symbolic operands to their equivalent machine addresses. • Decide the proper instruction format Convert the data constants to internal machine representations. • Generate relative address.



Proposition 2: statement of problems in assembler design. Concept structure

An assembler must do the following : 1. generate instructions:

i. Evaluate the mnemonic in the operation field to produce its machine code.

ii. Evaluate the subfield means find the value of each symbol, process literals , and assign addresses.

2. process pseudo ops : Example:

Inter mediate steps in assembling a program

During assembling the above program the symbols (FOUR,FIVE,TEMP) appear before they are defined, so it is convenient to make two passes for assembling. Pass 1: purpose – define symbols and literals. 1. determine length of machine instructions (MOTGET1) 2. keep track of location counter(LC). 3. remember values of symbols until pass 2 ( STSTO) 4. process some pseudo ops e.g. EQU, DS (POTGET1) 5. remember literals ( LITSTO) Pass 2 : purpose- generate object program 1. Look up value of symbols(STGET) 2. generating instructions (MOTGET2) 3. generate data (for DS,DC, and literals ) 4. process pseudo ops(POTGET2)

First pass Second pass Source program Relative

address Mnemonic Instruction

Relative address

Mnemonic Instruction

JOHN FOUR FIVE TEMP

START 0 USING *,15 ~ L 1, FIVE A 1, FOUR ST 1, TEMP DC F’4’ DC F’5’ DS 1F END

0 4 8 12 16 20

L 1,-(0,15) A 1,-(0,15) ST 1,-(0,15) 4 5 -

0 4 8 12 16 20

L 1,16(0,15) A 1,12(0,15) ST 1,20(0,15) 4 5 -



Proposition 3 : Data structures for assembler design. Data structures for pass1:

• Input source program

• Location counter(LC) to keep track of location of each instruction.

• Machine Operation Table(MOT) that consists symbolic mnemonic for each instruction along with its length.

• Pseudo Operation Table(POT) that indicates symbolic mnemonic and action to be taken for each pseudo op in pass1

• Symbol table(ST) that is used to store each label and its corresponding value.

• Literal table (LT) that is used to store each literal and its corresponding assigned location.

• A copy of input to be used later by pass 2 Data structures for pass2:

• A copy of source program input pass1

• Location counter(LC) to keep track of location of each instruction. • Machine Operation Table(MOT) that indicates for each instruction i) symbolic

mnemonic ii) length iii) binary machine opcode iv) format of instruction.

• Pseudo Operation Table(POT) that indicates symbolic mnemonic and action to be taken for each pseudo op in pass2

• Symbol table(ST) prepared by pass1,containing each label and its corresponding value.

• Base Table(BT) that indicates which registers are currently specified as base registers and their contents.

• Workspaces for a)holding various parts of instruction during assembling, b)producing a printed list

• Storage for holding the copy of assembled instructions in the format needed by the loader.

Proposition 4: Format of database in assembler design MOT: Format of machine operation table

Mnemonic Opcode (4 bytes) (characters)

Binary Opcode (1 byte) (hexadecimal)

Instruction length (2 bits) (binary)

Instruction format (3 bits) (binary)

“Abbbb” “AHbbb” . . . .

5A 4A . . . .

10 10 . . . .

001 001 . . . .

6 bytes per entry



POT: Format of pseudo operation table

ST: Format of symbol table BT: Format of machine operation table



A sample assembly program with illustration for symbol, literal, and base tables:

Sample assembly source program

Base table (showing only base registers in use)

1) After statement 2: Base Content 15 0 2) After statement 10: Base Content 15 6 3) After statement 12: Base Content 13 8064

15 6



Proposition 5 : use of one pass and two pass assemblers One-pass assemblers are used when

1. It is necessary or desirable to avoid a second pass over the source program 2. the external storage for the intermediate file between two passes is slow or is

inconvenient to use • Main problem: forward references to both data and instructions • One simple way to eliminate this problem: require that all areas be defined before they are referenced.

1. It is possible, although inconvenient, to do so for data items. 2. Forward jump to instruction items cannot be easily eliminated.

How to Handle Forward References Two pass assembler can resolve the forward references by using following method.

o Load-and-go assembler o Omits the operand address if the symbol has not yet been defined o Enters this undefined symbol into ST and indicates that it is undefined o Adds the address of this operand address to a list of forward references

associated with the ST entry o Scans the reference list and inserts the address when the definition for the

symbol is encountered. o Reports the error if there are still ST entries indicated undefined symbols at

the end of the program o Search ST for the symbol named in the END statement and jumps to this

location to begin execution if there is no error

4.0 Learning objectives : Intellectual skill

1. To understand design procedure of assembler 2. To understand data structures like LC,ST,BT,LT,MOT,POT etc. 3. To understand working sequence of assembler with another system programs loader, linker and compiler. Motor skills

1. Ability to design symbol table and literal table. 2. Ability to handle the entries in symbol table and literal table.

5.0 Stepwise procedure/ Algorithm: Algorithm : one pass

• Begin

• read first input line

• if OPCODE = ‘START’ then begin

• save #[Operand] as starting addr

• initialize LC to starting address

• write line to intermediate file

• read next line

• end( if START)

• else

• initialize LC to 0

• While OPCODE != ‘END’ do

• begin

• if this is not a comment line then

• beg

• if there is a symbol in the LABEL field then • begin

• search ST for LABEL

• if found then



• set error flag (duplicate symbol)

• else

• (if symbol) • search OPTAB for OPCODE

• if found then

• add 3 (instr length) to LC

• else if OPCODE = ‘WORD’ then

• add 3 to LC

• else if OPCODE = ‘RESW’ then

• add 3 * #[OPERAND] to LC

• else if OPCODE = ‘RESB’ then

• add #[OPERAND] to LC

• else if OPCODE = ‘BYTE’ then

• begin

• find length of constant in bytes

• add length to LC

• end

• else

• set error flag (invalid operation code)

• end (if not a comment)

• write line to intermediate file

• read next input line

• end { while not END}

• write last line to intermediate file

• Save (LC – starting address) as program length

• End {pass 1}

Algorithm : Two Pass

•

• Begin

• read 1st input line

• if OPCODE = ‘START’ then

• begin

• write listing line


• end • write Header record to object program

• initialize 1st Text record

• while OPCODE != ‘END’ do

• begin

• if this is not comment line then

• begin

• search OPTAB for OPCODE

• if found then

• begin

• if there is a symbol in OPERAND field then

• begin

• search ST for OPERAND field then

• if found then

• begin • store symbol value as operand address

• else



• begin

• store 0 as operand address

• set error flag (undefined symbol) • end

• end (if symbol)

• else store 0 as operand address

• assemble the object code instruction

• else if OPCODE = ‘BYTE’ or ‘WORD” then

• convert constant to object code

• if object code doesn’t fit into current Text record then

• begin

• Write text record to object code

• initialize new Text record

• end

• add object code to Text record

• end {if not comment}



• end



• write last listing line

• End {Pass 2}

6.0 Conclusion: 1. The design of assembler requires to follow the steps

1. Specify the problem 2. ………………………… 3. Define format of data structures 4. ………………………… 5…………………………

6. Repeat 1 through 5 on modules

2. While assembler design two passes are needed due to …………………………………………………………….. ………

……………………………………………………………..………

7.0 Laboratory practice:

1) For the following assembly code, design a)symbol table b)literal table

c)base table SIMPLE START BALR 15, 0 USING *, 15 LOOP L R1, TWO A R2, TWO ST R1, FOUR CL1 FOUR +3,4 BNE LOOP BR 14 R1 EQU 1 TWO DC F’2’ FOUR DS F END



2) For a given a data structure MOT.Write a “C” program to enter mnemonic opcode in to MOT and search the particular mnemonic opcode form MOT.

6 bytes per entry

Mnemonic Opcode (4 bytes) (characters)

Binary Opcode (1 byte) (hexadecimal)

Instruction length (2 bits) (binary)

Instruction format (3 bits) (binary)

“Abbbb” “AHbbb” . . . .

5A 4A . . . .

10 10 . . . .

001 001 . . . .

7.0 Questions:

Write answers to Q.....Q.....Q.....Q.....and Q..... (Teacher shall allot the questions) 1. What do you mean by assembler. 2. List the assembler design steps. 3. State the use of following data structures. a) Symbol Table b) Literal Table. c) Base table. d) Location Counter . e) MOT f) POT

4.Draw the flowchart for Pass1 of assembler design. 5.Draw the flowchart for Pass2 of assembler design.

(Space for answers)



(Space for answers)

System Programing Lab

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